SOLUTION: Ed has 95 cents in dimes and nickels. The total number of coins is one more than twice the number of dimes. How many dimes does she have?
I'm confused on this because it doesn't d
Algebra ->
Customizable Word Problem Solvers
-> Coins
-> SOLUTION: Ed has 95 cents in dimes and nickels. The total number of coins is one more than twice the number of dimes. How many dimes does she have?
I'm confused on this because it doesn't d
Log On
Question 1016261: Ed has 95 cents in dimes and nickels. The total number of coins is one more than twice the number of dimes. How many dimes does she have?
I'm confused on this because it doesn't doesn't tell me a total number so I'm
Not sure what to go off of
The only equation I could think of was
.10d + .05n = .95
Not the other one we should have if someone could help
Me find the other equation and find the amount of coins.
Please and thank you ! Found 2 solutions by Alan3354, ankor@dixie-net.com:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Ed has 95 cents in dimes and nickels. The total number of coins is one more than twice the number of dimes. How many dimes does she have?
=======================
5n + 10d = 95
n+d = 2d+1
You can put this solution on YOUR website! Ed has 95 cents in dimes and nickels. The total number of coins is one more than twice the number of dimes. How many dimes does she have?
I'm confused on this because it doesn't doesn't tell me a total number so I'm
Not sure what to go off of
The only equation I could think of was
.10d + .05n = .95
:
Write an equation for the statement
"the total number of coins is one more than twice the number of dimes."
d + n = 2d + 1
n = 2d - d + 1
n = d + 1
now, in your first equation you can substitute (d+1) for n
.10d + .05(d+1) = .95
.10d + .05d + .05 = .95
.15d = .95 - .05
.15d = .90
d = .90/.15
d = 6 dimes
I'll let you find the no. of nickels, check your solutions in the 1st equation
